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IrinaK [193]
3 years ago
13

The gravitational force between two asteroids is 1,000,000 n. what will the force be if the distance between the asteroids is do

ubled?
Physics
1 answer:
Alina [70]3 years ago
3 0

To solve this problem, we use the formula:

F = G m1 m2 / r^2

where F is gravitational force, G is constant, m1 and m2 are masses while r is the distance between the two asteroids

Since G m1 m2 is constant, therefore:

F1 r1^2 = F2 r2^2

So if r2 = 2 r1:

(1,000,000 N) (r1^2) = F2 * (2 r1)^2

<span>F2 = 250,000 N</span>

<span>
</span>

<span>It was divided by 4</span>

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What will the stopping distance be for a 3,000-kg car if -3,000 N of force are applied when the car is traveling 10 m/s?
Harlamova29_29 [7]

F = force applied to stop the car = - 3000 N

m = mass of the car = 3000 kg

a = acceleration of the car = ?

v₀ = initial velocity of the car before the force is applied to stop it = 10 m/s

v = final velocity of the car when it comes to stop = 0 m/s

d = stopping distance of the car

acceleration of the car is given as

a = F/m

inserting the values

a = - 3000/3000

a = - 1 m/s²

using the kinematics equation

v² = v²₀ + 2 a d

inserting the values

0² = 10² + 2 (-1) d

0 = 100 - 2 d

2 d = 100

d = 100/2

d = 50 m


hence the correct choice is

C. 50 m

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look if its at school

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Which note-taking method did Samantha use?
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D) Cornell method

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CORRECT ON E2020

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What an object at rest stays at rest and an object in motion with the same speed and in the same direction unless acted upon by
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True.

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A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill. there is no friction, what is
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Answer:

the maximum vertical height the person in the cart can reach is 18.42 m

Explanation:

Given;

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mass of the cart, m₂ = 43 kg

acceleration due to gravity, g = 9.8 m/s²

final speed of the cart before it goes up the hill, v = 19 m/s

Apply the principle of conservation of energy;

mgh_{max} = \frac{1}{2}mv^2_{max}\\\\ gh_{max} = \frac{1}{2}v^2_{max}\\\\h_{max} = \frac{v^2_{max}}{2g} \\\\h_{max} =\frac{(19)^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m

Therefore, the maximum vertical height the person in the cart can reach is 18.42 m

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